Sql 或中的x构成： SELECT id, name, other_columns FROM messages WHERE id IN ( SELECT MAX(id) FROM messages GROUP BY name );_Sql_Mysql_Group By_Greatest N Per Group

Sql 或中的x构成： SELECT id, name, other_columns FROM messages WHERE id IN ( SELECT MAX(id) FROM messages GROUP BY name );

sql mysql

Sql 或中的x构成： SELECT id, name, other_columns FROM messages WHERE id IN ( SELECT MAX(id) FROM messages GROUP BY name );,sql,mysql,group-by,greatest-n-per-group,Sql,Mysql,Group By,Greatest N Per Group,我不知道与其他一些解决方案相比，它的性能如何，但它在我的300多万行的表中运行得非常出色。（4秒执行，1200多个结果）这应该可以在MySQL和SQL Server上使用。我还没有使用大型数据库进行测试，但我认为这可能比连接表更快： SELECT *, Max(Id) FROM messages GROUP BY Name CREATE TABLE temperature( id INT UNSIGNED NOT NULL AUTO_INCREMENT, groupID INT U

我不知道与其他一些解决方案相比，它的性能如何，但它在我的300多万行的表中运行得非常出色。（4秒执行，1200多个结果）

这应该可以在MySQL和SQL Server上使用。

我还没有使用大型数据库进行测试，但我认为这可能比连接表更快：

SELECT *, Max(Id) FROM messages GROUP BY Name

CREATE TABLE temperature(
  id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
  groupID INT UNSIGNED NOT NULL, 
  recordedTimestamp TIMESTAMP NOT NULL, 
  recordedValue INT NOT NULL,
  INDEX groupIndex(groupID, recordedTimestamp), 
  PRIMARY KEY (id)
);

CREATE TEMPORARY TABLE selected_group(id INT UNSIGNED NOT NULL, PRIMARY KEY(id));

子查询解决方案

按连接条件求解

从消息m1中选择m1.*
左外连接消息m2
在（m1.id上，这里是另一种获取最后一条相关记录的方法，使用带有order by的GROUP_CONCAT
和SUBSTRING_INDEX
从列表中选择一条记录
SELECT 
  `Id`,
  `Name`,
  SUBSTRING_INDEX(
    GROUP_CONCAT(
      `Other_Columns` 
      ORDER BY `Id` DESC 
      SEPARATOR '||'
    ),
    '||',
    1
  ) Other_Columns 
FROM
  messages 
GROUP BY `Name` 

上面的查询将把所有其他_列
分组，这些列位于相同的名称
组中，并使用按id DESC排序
将所有其他_列
按降序与提供的分隔符连接在一个特定的组中，我在这个lis上使用了|
，使用了子字符串
t将选择第一个
Hi@Vijay Dev如果您的表消息包含自动递增主键的Id，则要根据主键获取最新记录，您的查询应如下所示：
SELECT m1.* FROM messages m1 INNER JOIN (SELECT max(Id) as lastmsgId FROM messages GROUP BY Name) m2 ON m1.Id=m2.lastmsgId

你也可以从这里观看

第一个解决方案
SELECT d1.ID,Name,City FROM Demo_User d1
INNER JOIN
(SELECT MAX(ID) AS ID FROM Demo_User GROUP By NAME) AS P ON (d1.ID=P.ID);

第二种解决方案
SELECT * FROM (SELECT * FROM Demo_User ORDER BY ID DESC) AS T GROUP BY NAME ;

如果要为每个名称
指定最后一行，则可以按名称
为每个行组指定行号，并按Id
降序排列
查询
SELECT t1.Id, 
       t1.Name, 
       t1.Other_Columns
FROM 
(
     SELECT Id, 
            Name, 
            Other_Columns,
    (
        CASE Name WHEN @curA 
        THEN @curRow := @curRow + 1 
        ELSE @curRow := 1 AND @curA := Name END 
    ) + 1 AS rn 
    FROM messages t, 
    (SELECT @curRow := 0, @curA := '') r 
    ORDER BY Name,Id DESC 
)t1
WHERE t1.rn = 1
ORDER BY t1.Id;

这个怎么样：
SELECT DISTINCT ON (name) *
FROM messages
ORDER BY name, id DESC;

我也有类似的问题（在postgresql上），在1M记录表上。这个解决方案需要1.7秒，而左连接的需要44秒。
在我的例子中，我必须针对空值过滤your name字段的Correspondant，从而使性能提高0.2秒，这是我的解决方案：
DROP TABLE IF EXISTS UniqueIDs; 
CREATE Temporary table UniqueIDs (id Int(11)); 

INSERT INTO UniqueIDs 
    (SELECT T1.ID FROM Table T1 LEFT JOIN Table T2 ON 
    (T1.Field1 = T2.Field1 AND T1.Field2 = T2.Field2 #Comparison Fields  
    AND T1.ID < T2.ID) 
    WHERE T2.ID IS NULL); 

DELETE FROM Table WHERE id NOT IN (SELECT ID FROM UniqueIDs);

SELECT 
  DISTINCT NAME,
  MAX(MESSAGES) OVER(PARTITION BY NAME) MESSAGES 
FROM MESSAGE;

速度相当快的方法如下
SELECT * 
FROM messages a
WHERE Id = (SELECT MAX(Id) FROM messages WHERE a.Name = Name)

结果
Id  Name    Other_Columns
3   A   A_data_3
5   B   B_data_2
6   C   C_data_1

很明显，有很多不同的方法可以获得相同的结果，你的问题似乎是，在MySQL中，获得每个组最后结果的有效方法是什么。如果你正在处理大量数据，并且假设你使用的是InnoDB，甚至是最新版本的MySQL（如5.7.21和8.0.4-rc）那么可能没有一个有效的方法来做到这一点
我们有时需要对超过6000万行的表执行此操作
对于这些示例，我将使用只有大约150万行的数据，其中查询需要查找数据中所有组的结果。在我们的实际案例中，我们通常需要返回大约2000个组的数据（假设不需要检查太多数据）
我将使用以下表格：
SELECT *, Max(Id) FROM messages GROUP BY Name

CREATE TABLE temperature(
  id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
  groupID INT UNSIGNED NOT NULL, 
  recordedTimestamp TIMESTAMP NOT NULL, 
  recordedValue INT NOT NULL,
  INDEX groupIndex(groupID, recordedTimestamp), 
  PRIMARY KEY (id)
);

CREATE TEMPORARY TABLE selected_group(id INT UNSIGNED NOT NULL, PRIMARY KEY(id)); 

温度表包含大约150万条随机记录，以及100个不同的组。
选定的_组由这100个组填充（在我们的案例中，所有组的填充率通常小于20%）
由于此数据是随机的，这意味着多行可以具有相同的recordedTimestamp。我们想要的是按照groupID的顺序获取所有选定组的列表，每个组具有最后一个recordedTimestamp，如果同一组具有多个类似的匹配行，则获取这些行的最后一个匹配id
如果假设MySQL有一个last（）函数，它以特殊的ORDER BY子句返回最后一行的值，那么我们可以简单地执行以下操作：
SELECT 
  last(t1.id) AS id, 
  t1.groupID, 
  last(t1.recordedTimestamp) AS recordedTimestamp, 
  last(t1.recordedValue) AS recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.groupID = g.id
ORDER BY t1.recordedTimestamp, t1.id
GROUP BY t1.groupID;

在这种情况下，它只需要检查几百行，因为它不使用任何正常的GROUP BY函数。这将在0秒内执行，因此非常高效。
请注意，通常在MySQL中，我们会在GROUP BY子句之后看到ORDER BY子句，但此ORDER BY子句用于确定last（）函数的顺序，如果它在GROUP BY之后，则它将对组进行排序。如果不存在GROUP BY子句，则所有返回行中的最后值都将相同
然而MySQL并没有这样的功能，所以让我们看看它有什么不同的想法，并证明这些都不是有效的
示例1
SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.id = (
  SELECT t2.id
  FROM temperature t2 
  WHERE t2.groupID = g.id
  ORDER BY t2.recordedTimestamp DESC, t2.id DESC
  LIMIT 1
);

这检查了3009254行，在5.7.21上花费了约0.859秒，在8.0.4-rc上花费了稍长的时间
示例2
SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM temperature t1
INNER JOIN ( 
  SELECT max(t2.id) AS id   
  FROM temperature t2
  INNER JOIN (
    SELECT t3.groupID, max(t3.recordedTimestamp) AS recordedTimestamp
    FROM selected_group g
    INNER JOIN temperature t3 ON t3.groupID = g.id
    GROUP BY t3.groupID
  ) t4 ON t4.groupID = t2.groupID AND t4.recordedTimestamp = t2.recordedTimestamp
  GROUP BY t2.groupID
) t5 ON t5.id = t1.id;

这检查了1505331行，在5.7.21上花费了约1.25秒，在8.0.4-rc上花费了稍长的时间
示例3
SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM temperature t1
WHERE t1.id IN ( 
  SELECT max(t2.id) AS id   
  FROM temperature t2
  INNER JOIN (
    SELECT t3.groupID, max(t3.recordedTimestamp) AS recordedTimestamp
    FROM selected_group g
    INNER JOIN temperature t3 ON t3.groupID = g.id
    GROUP BY t3.groupID
  ) t4 ON t4.groupID = t2.groupID AND t4.recordedTimestamp = t2.recordedTimestamp
  GROUP BY t2.groupID
)
ORDER BY t1.groupID;

这检查了3009685行，在5.7.21上花费了约1.95秒，在8.0.4-rc上花费了稍长的时间
示例4
SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.id = (
  SELECT max(t2.id)
  FROM temperature t2 
  WHERE t2.groupID = g.id AND t2.recordedTimestamp = (
      SELECT max(t3.recordedTimestamp)
      FROM temperature t3 
      WHERE t3.groupID = g.id
    )
);

这检查了6137810行，在5.7.21上花费了约2.2秒，在8.0.4-rc上花费了稍长的时间
示例5
SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM (
  SELECT 
    t2.id, 
    t2.groupID, 
    t2.recordedTimestamp, 
    t2.recordedValue, 
    row_number() OVER (
      PARTITION BY t2.groupID ORDER BY t2.recordedTimestamp DESC, t2.id DESC
    ) AS rowNumber
  FROM selected_group g 
  INNER JOIN temperature t2 ON t2.groupID = g.id
) t1 WHERE t1.rowNumber = 1;

这检查了6017808行，在8.0.4-rc上耗时约4.2秒
示例6
SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM (
  SELECT 
    last_value(t2.id) OVER w AS id, 
    t2.groupID, 
    last_value(t2.recordedTimestamp) OVER w AS recordedTimestamp, 
    last_value(t2.recordedValue) OVER w AS recordedValue
  FROM selected_group g
  INNER JOIN temperature t2 ON t2.groupID = g.id
  WINDOW w AS (
    PARTITION BY t2.groupID 
    ORDER BY t2.recordedTimestamp, t2.id 
    RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
  )
) t1
GROUP BY t1.groupID;

这检查了6017908行，在8.0.4-rc上花费了约17.5秒
示例7
SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM selected_group g
INNER JOIN temperature t1 ON t1.groupID = g.id
LEFT JOIN temperature t2 
  ON t2.groupID = g.id 
  AND (
    t2.recordedTimestamp > t1.recordedTimestamp 
    OR (t2.recordedTimestamp = t1.recordedTimestamp AND t2.id > t1.id)
  )
WHERE t2.id IS NULL
ORDER BY t1.groupID;

这一列花了很长时间，所以我不得不终止它。
如果性能确实是您关心的问题，您可以在表中引入一个新列，名为BIT类型的IsLastInGroup

对于最后一列，将其设置为true，并在每一行插入/更新/删除时维护它。写入速度会慢一些，但您会从读取中受益。这取决于您的用例，我建议您仅在以读取为中心的情况下使用
因此，您的查询将如下所示：
SELECT * FROM Messages WHERE IsLastInGroup = 1

您可以通过计数进行分组，还可以获得组的最后一项，如：
SELECT 
    user,
    COUNT(user) AS count,
    MAX(id) as last
FROM request 
GROUP BY user

我们将研究如何使用MySQL获取一组记录中的最后一条记录
id类别\u id帖子\u标题
1标题1

2 1标题2

3 1标题3

42标题4

52
select m1.* from messages m1 
left outer join messages m2 
on ( m1.id<m2.id and m1.name=m2.name )
where m2.id is null

SELECT 
  `Id`,
  `Name`,
  SUBSTRING_INDEX(
    GROUP_CONCAT(
      `Other_Columns` 
      ORDER BY `Id` DESC 
      SEPARATOR '||'
    ),
    '||',
    1
  ) Other_Columns 
FROM
  messages 
GROUP BY `Name` 

SELECT 
  column1,
  column2 
FROM
  table_name 
WHERE id IN 
  (SELECT 
    MAX(id) 
  FROM
    table_name 
  GROUP BY column1) 
ORDER BY column1 ;

SELECT m1.* FROM messages m1 INNER JOIN (SELECT max(Id) as lastmsgId FROM messages GROUP BY Name) m2 ON m1.Id=m2.lastmsgId

SELECT d1.ID,Name,City FROM Demo_User d1
INNER JOIN
(SELECT MAX(ID) AS ID FROM Demo_User GROUP By NAME) AS P ON (d1.ID=P.ID);

SELECT * FROM (SELECT * FROM Demo_User ORDER BY ID DESC) AS T GROUP BY NAME ;

SELECT t1.Id, 
       t1.Name, 
       t1.Other_Columns
FROM 
(
     SELECT Id, 
            Name, 
            Other_Columns,
    (
        CASE Name WHEN @curA 
        THEN @curRow := @curRow + 1 
        ELSE @curRow := 1 AND @curA := Name END 
    ) + 1 AS rn 
    FROM messages t, 
    (SELECT @curRow := 0, @curA := '') r 
    ORDER BY Name,Id DESC 
)t1
WHERE t1.rn = 1
ORDER BY t1.Id;

select * from messages group by name desc

SELECT DISTINCT ON (name) *
FROM messages
ORDER BY name, id DESC;

SELECT 
  DISTINCT NAME,
  MAX(MESSAGES) OVER(PARTITION BY NAME) MESSAGES 
FROM MESSAGE;

SELECT * 
FROM messages a
WHERE Id = (SELECT MAX(Id) FROM messages WHERE a.Name = Name)

Id  Name    Other_Columns
3   A   A_data_3
5   B   B_data_2
6   C   C_data_1

CREATE TABLE temperature(
  id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
  groupID INT UNSIGNED NOT NULL, 
  recordedTimestamp TIMESTAMP NOT NULL, 
  recordedValue INT NOT NULL,
  INDEX groupIndex(groupID, recordedTimestamp), 
  PRIMARY KEY (id)
);

CREATE TEMPORARY TABLE selected_group(id INT UNSIGNED NOT NULL, PRIMARY KEY(id)); 

SELECT 
  last(t1.id) AS id, 
  t1.groupID, 
  last(t1.recordedTimestamp) AS recordedTimestamp, 
  last(t1.recordedValue) AS recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.groupID = g.id
ORDER BY t1.recordedTimestamp, t1.id
GROUP BY t1.groupID;

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.id = (
  SELECT t2.id
  FROM temperature t2 
  WHERE t2.groupID = g.id
  ORDER BY t2.recordedTimestamp DESC, t2.id DESC
  LIMIT 1
);

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM temperature t1
INNER JOIN ( 
  SELECT max(t2.id) AS id   
  FROM temperature t2
  INNER JOIN (
    SELECT t3.groupID, max(t3.recordedTimestamp) AS recordedTimestamp
    FROM selected_group g
    INNER JOIN temperature t3 ON t3.groupID = g.id
    GROUP BY t3.groupID
  ) t4 ON t4.groupID = t2.groupID AND t4.recordedTimestamp = t2.recordedTimestamp
  GROUP BY t2.groupID
) t5 ON t5.id = t1.id;

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM temperature t1
WHERE t1.id IN ( 
  SELECT max(t2.id) AS id   
  FROM temperature t2
  INNER JOIN (
    SELECT t3.groupID, max(t3.recordedTimestamp) AS recordedTimestamp
    FROM selected_group g
    INNER JOIN temperature t3 ON t3.groupID = g.id
    GROUP BY t3.groupID
  ) t4 ON t4.groupID = t2.groupID AND t4.recordedTimestamp = t2.recordedTimestamp
  GROUP BY t2.groupID
)
ORDER BY t1.groupID;

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.id = (
  SELECT max(t2.id)
  FROM temperature t2 
  WHERE t2.groupID = g.id AND t2.recordedTimestamp = (
      SELECT max(t3.recordedTimestamp)
      FROM temperature t3 
      WHERE t3.groupID = g.id
    )
);

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM (
  SELECT 
    t2.id, 
    t2.groupID, 
    t2.recordedTimestamp, 
    t2.recordedValue, 
    row_number() OVER (
      PARTITION BY t2.groupID ORDER BY t2.recordedTimestamp DESC, t2.id DESC
    ) AS rowNumber
  FROM selected_group g 
  INNER JOIN temperature t2 ON t2.groupID = g.id
) t1 WHERE t1.rowNumber = 1;

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM (
  SELECT 
    last_value(t2.id) OVER w AS id, 
    t2.groupID, 
    last_value(t2.recordedTimestamp) OVER w AS recordedTimestamp, 
    last_value(t2.recordedValue) OVER w AS recordedValue
  FROM selected_group g
  INNER JOIN temperature t2 ON t2.groupID = g.id
  WINDOW w AS (
    PARTITION BY t2.groupID 
    ORDER BY t2.recordedTimestamp, t2.id 
    RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
  )
) t1
GROUP BY t1.groupID;

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM selected_group g
INNER JOIN temperature t1 ON t1.groupID = g.id
LEFT JOIN temperature t2 
  ON t2.groupID = g.id 
  AND (
    t2.recordedTimestamp > t1.recordedTimestamp 
    OR (t2.recordedTimestamp = t1.recordedTimestamp AND t2.id > t1.id)
  )
WHERE t2.id IS NULL
ORDER BY t1.groupID;

SELECT * FROM Messages WHERE IsLastInGroup = 1

SELECT * FROM table_name WHERE primary_key IN (SELECT MAX(primary_key) FROM table_name GROUP BY column_name )

SELECT 
    user,
    COUNT(user) AS count,
    MAX(id) as last
FROM request 
GROUP BY user

SELECT 
  *
FROM 
  message 

WHERE 
  `Id` IN (
    SELECT 
      MAX(`Id`) 
    FROM 
      message 
    GROUP BY 
      `Name`
  ) 
ORDER BY 
   `Id` DESC

select * from properties p
join (
    select max(m2_price) as max_price 
    from properties 
    group by program_id
) p2 on (p.program_id = p2.program_id)
having p.m2_price = max_price

WITH Temp_table AS
(
    Select id, name, othercolumns, ROW_NUMBER() over (PARTITION BY name ORDER BY ID 
    desc)as rank from messages
)
Select id, name,othercolumns from Temp_table where rank=1

select * from `data` where `id` in (select max(`id`) from `data` group by `name_id`)

select *, max(id) from messages group by name 

SELECT GROUP_CONCAT(id ORDER BY id DESC LIMIT 1) AS id,
       name,
       GROUP_CONCAT(Other_columns ORDER BY id DESC LIMIT 1) AS Other_columns
FROM t
GROUP BY name;